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Thread: Second derivatives

  1. #1
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    Exclamation Second derivatives

    I was searching the board and found a couple posts on this but they didn't really fit the format I was doing derivatives in. $\displaystyle F(x)=-1/(x^2+5)$ is my function. For the first derivative I have $\displaystyle F(x)=2x/(x^2+5)^2$. For some reason, all of my answers I get for the second derivative are wrong. Can anyone show me the steps you take to get to it?
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  2. #2
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    $\displaystyle f(x) = -(x^2+5)^{-1}$

    $\displaystyle f'(x) = 2x(x^2+5)^{-2}$

    $\displaystyle f''(x) = 2x \cdot [-2(x^2+5)^{-3} \cdot 2x] + (x^2+5)^{-2} \cdot 2$

    $\displaystyle f''(x) = \frac{-8x^2}{(x^2+5)^3} + \frac{2}{(x^2+5)^2}$

    $\displaystyle f''(x) = \frac{-8x^2}{(x^2+5)^3} + \frac{2(x^2+5)}{(x^2+5)^3}$

    $\displaystyle f''(x) = \frac{2(5 - 3x^2)}{(x^2+5)^3}$
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by dougmoser View Post
    I was searching the board and found a couple posts on this but they didn't really fit the format I was doing derivatives in. $\displaystyle F(x)=-1/(x^2+5)$ is my function. For the first derivative I have $\displaystyle F(x)=2x/(x^2+5)^2$. For some reason, all of my answers I get for the second derivative are wrong. Can anyone show me the steps you take to get to it?
    Set
    $\displaystyle F(x):=-\frac{1}{x^{2}+5}.$
    Then
    $\displaystyle F^{\prime}(x)=\frac{2x}{(x^{2}+5)^{2}}.$
    And then
    $\displaystyle F^{\prime\prime}(x)=\frac{(2x)^{\prime}(x^{2}+5)^{ 2}-(2x)\big((x^{2}+5)^{2}\big)^{\prime}}{\big((x^{2}+ 5)^{2}\big)^{2}}$ (Chain rule)
    .........$\displaystyle =\frac{2(x^{2}+5)^{2}-(2x)\big(2(x^{2}+5)^{\prime}\big)}{(x^{2}+5)^{4}}$ (Chain rule)
    .........$\displaystyle =\frac{2(x^{2}+5)^{2}-(2x)\big(2(2x)(x^{2}+5)\big)}{(x^{2}+5)^{4}}$
    .........$\displaystyle =-\frac{6x^{4}+20x^{2}-50}{(x^{2}+5)^{4}}$
    .........$\displaystyle =-\frac{2(3x^{2}-5)(x^{2}+5)}{(x^{2}+5)^{4}}$
    .........$\displaystyle =-\frac{2(3x^{2}-5)}{(x^{2}+5)^{3}}.$
    Last edited by bkarpuz; Oct 3rd 2008 at 01:57 PM. Reason: '(Chain rule)' is added.
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  4. #4
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    Thanks a lot guys!
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